I was wondering how to prove this
Problem:
Let m be a positive integer and let n be an integer obtained from m by rearranging the digits of m in some way. (for example 381762 reaaranging 378126) prove m-n is divisble 9

thanks :D

August 14th 2011, 01:45 AM

pickslides

Re: Divisible by 9

If the digits of any number sum to a multiple of nine then it is also divisible by nine. Rearranging them won't change this.
(Smirk)

August 14th 2011, 01:46 AM

Moo

Re: Divisible by 9

Hello,

Well a simple way to do it is to remember that, S(.) being the sum of the digits of a number, (this can be easily proved with congruences, writing the number as powers of 10). And since (rearranging the digits won't change their sum !), then

August 14th 2011, 01:46 AM

Moo

Re: Divisible by 9

Quote:

Originally Posted by pickslides

If the digits of any number sum to a multiple of nine then it is also divisible by nine. Rearranging them won't change this.
(Smirk)

Yes, but it isn't stated that m is divisible by 9 (Smirk)

August 14th 2011, 01:48 AM

thimellecx

Re: Divisible by 9

what if the sum of its digits is not multiple to nine like 721 and 172

August 14th 2011, 01:57 AM

pickslides

Re: Divisible by 9

Quote:

Originally Posted by Moo

Yes, but it isn't stated that m is divisible by 9 (Smirk)

Well I did say if..
(Rofl)

August 14th 2011, 08:13 AM

HallsofIvy

Re: Divisible by 9

More generally, if the remainder of the sum of digits of a number, divided by 9, is n, then the remainder of the number, divided by 9, is also n. Since rearranging the digits in a number does not change the sum of digits, it also does not change the remainder when divided by n. Let A be the original number, and let n be its remainder when divided by 9: A= 9k+ n for some integer k. If B is any rearrangement of the digits of A, then B= 9j+ n for some integer j. A-B= (9k+ n)- (9j+ n)= 9(k- j).

You can prove that statement about the remainders by looking at the numbers expansion in powers of 10:

August 14th 2011, 10:17 AM

Moo

Re: Divisible by 9

Quote:

Originally Posted by HallsofIvy

More generally, if the remainder of the sum of digits of a number, divided by 9, is n, then the remainder the number, divided by 9, is also n. Since rearranging the digits in a number does not change the sum of digits, it also does not change the remainder when divided by n. Let A be the original number, and let n be its remainder when divided by 9: A= 9k+ n for some integer k. If B is any rearrangement of the digits of A, then B= 9j+ n for some integer j. A-B= (9k+ n)- (9j+ n)= 9(k- j).

You can prove that statement about the remainders by looking at the numbers expansion in powers of 10:

Hm well, isn't that equivalent to what I wrote in #3 ? (Itwasntme) (I'm quite disturbed by your "more generally", not by the fact that you're giving another way to see through the answer)
The congruence gives the remainder, so...

And the OP seems to have understood the solution, given the private messages he's sent me...

August 14th 2011, 10:49 AM

HallsofIvy

Re: Divisible by 9

Yes, you are completely correct! I just did not read your post properly. The "more generally" was referring to PickSlides' post #2.

August 14th 2011, 08:26 PM

SammyS

Re: Divisible by 9

Quote:

Originally Posted by Moo

Hm well, isn't that equivalent to what I wrote in #3 ? (Itwasntme) (I'm quite disturbed by your "more generally", not by the fact that you're giving another way to see through the answer)
The congruence gives the remainder, so...

And the OP seems to have understood the solution, given the private messages he's sent me...

Yes, Moo. Your post, #3, was ignored earlier, particularly by OP.

August 15th 2011, 01:12 AM

Moo

Re: Divisible by 9

Quote:

Originally Posted by SammyS

Yes, Moo. Your post, #3, was ignored earlier, particularly by OP.

Yes, I'm very sad that it was ignored !! (Rofl)

And the OP has thanked it and sent me a pm about what I wrote in it, so I guess he didn't ignore it !

August 17th 2011, 03:48 AM

livingdream

Re: Divisible by 9

well sum adds up to 27 so the number is divisble by 9 and 3 and even if you rearrange it adds to same